Divide with remainder $\frac{x^2}{x^2 + x + 2}$ I am having a hard time long dividing: 
$$\frac{x^2}{x^2 + x + 2}.$$
Could someone please show a step by step way to divide this, as I can only get it down to : $1 + \frac{x^2}{x + 2}$.
Thank you for your time!
 A: You can do long division of polynomials exactly like one does long division of integers on paper. 
Set it:
                       ___________________________
         x^2+x+2      |   x^2

Now: look at the leading term of the divisor: $x^2$; the leading term of the dividend: $x^2$; how many times does the leading term of the divisor go into the leading term of the dividend? $x^2$ goes into $x^2$ once. So we put that as our first "digit" of the quotient:
                            1
                       _______________________________                 
        x^2+x+2       |    x^2

Now, multiply $1$ by $x^2+x+2$, and subtract it from the dividend
                            1
                       _______________________________                
        x^2+x+2       |    x^2
                        - (x^2 + x + 2)
                        ---------------
                               - x - 2

Now, look at the leading term of what's left: $-x$; and the leading term of the divisor: $x^2$. How many times does $x^2$ go into $-x$? Zero times. So we're done with the division: the quotient is $1$, the remainder is $-x-2$. So we have that
$$\frac{x^2}{x^2+x+2} = 1 + \frac{-x-2}{x^2+x+2} = 1 - \frac{x+2}{x^2+x+2}.$$
A: Since $x^2 - 1 \cdot (x^2 + x + 2) = -x - 2$ (one step of polynomial division), we get $x^2 : (x^2 + x + 2) = 1$, remainder $-x-2$, or equivalently, $$\frac{x^2}{x^2 + x + 2} = 1 + \frac{-x-2}{x^2 + x + 2}.$$ This is, in fact, the final result, since the degree of $-x-2$ is less than that of $x^2 + x + 2$.
A: Hint: try expressing the fraction as
$$\frac{x^2}{x^2+x+2} = \frac{\quad x^2\quad+(x+2)\quad-(x+2)\quad}{x^2+x+2}.$$
